Kinetic Energy Poll Ball Question

1. The problem statement, all variables and given/known data
Two identical billiard balls are at rest on a frictionless, level surface, touching each other at one common point. A third, identical ball, the cue ball, is approaching along the common tangent with a constant speed of 20m/s. Assuming a completely elastic collision, with no spin on any balls, and making(reasonable) assumptions about symetry, calculate the velocity of the cue ball after the collision. The graph is as shown below

(the masses of the 3 balls are the same, but the value is unknown)
o------->8
20m/s

2. Relevant equations
I think two equations could be used in this question, one is
Pt = (20m/s)M = Pt`= (v1 m/s)M + (v2 m/s)M + (v3 m/s)M
the other one is The Conservation of Energy
Ek= Ek`

(M(20M/S)^2)/2 = (M(v1 m/s)^2)/2 + (M(v2 m/s)^2)/2 + (M(v3 m/s)^2)/2

3. The attempt at a solution
The only thing i can do is assume that ball1 and ball2 have the same speed after the collision
so that i got 2 equations

no.1 20M= 2(V1)M + (Vc)M, where V1 is the velocity of ball1 and ball2,Vc is the value we r looking for

no.2 (20^2)M/2 = 2M(V1^2)/2 + (Vc^2)M/2

and by substituting 20M-2V1=Vc in no.1 into no.2, I got V1 = 13.33, which will result in Vc= - 6.67m/s

but the answer in the book says Vc=-4m/s ... Did i do something wrong with my equations? Helps/Suggestions are really appreciated, Thanks in advanced!!

[QUOTE=lilbunnyf;1998392]1. 2. Relevant equations
I think two equations could be used in this question, one is
Pt = (20m/s)M = Pt`= (v1 m/s)M + (v2 m/s)M + (v3 m/s)M
the other one is The Conservation of Energy
Ek= Ek`

(M(20M/S)^2)/2 = (M(v1 m/s)^2)/2 + (M(v2 m/s)^2)/2 + (M(v3 m/s)^2)/2

QUOTE]

V1, V2 and V3 are not in the same direction. Take the camponents of V2 and V3 in the direction and perpendicular to the direction of cue ball.

[QUOTE=lilbunnyf;1998392]1. 2. Relevant equations
I think two equations could be used in this question, one is
Pt = (20m/s)M = Pt`= (v1 m/s)M + (v2 m/s)M + (v3 m/s)M
the other one is The Conservation of Energy
Ek= Ek`

(M(20M/S)^2)/2 = (M(v1 m/s)^2)/2 + (M(v2 m/s)^2)/2 + (M(v3 m/s)^2)/2

QUOTE]

V1, V2 and V3 are not in the same direction. Take the camponents of V2 and V3 in the direction and perpendicular to the direction of cue ball.

V2 and V3 are in opposite direction since they are perpendicular to V1? Btw why would they be perpendicular?

No he's not saying they're perpendicular, he's saying that v2 and v3 are vectors with some general directions (after the collisions) that you don't know, and therefore you're going to have to break up these vectors into *components* parallel to, and perpendicular to, the cue ball's initial direction.